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andriy [413]
2 years ago
13

Find the equation, in vertex form, for a parabola with a vertex of (2,-5) that goes through the point (3,4).

Mathematics
1 answer:
FromTheMoon [43]2 years ago
7 0
Y=9(x−2)²<span>−5 is the solution
</span>
Explanation in full below

Using the vertex co-ordinates we can deduct the formula
y=a(x-2)²-5 

using the co-ordinate (3,4), use x=3, y=4 to find a

y=a(x-2)²-5 
4=a(3-2)² -5
4=a(1)²-5
4=a-5
a=9

Subsituting a into the first equation
y=a(x-2)²-5 
y=9(x-2)²-5 

The solution is y=9(x-2)²-5 
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Z3=-64 help please I will give brainliest
snow_lady [41]

Answer:

-4

Step-by-step explanation:

<h2><u>An exponent is multiplying a number by itself as many times as the exponent says so.</u></h2><h2><u /></h2><h2><u>So we need to find a number that times itself 3 times.</u></h2>

-4 fits this.

-4 x -4 x -4

==> -64

z = -4

6 0
3 years ago
Read 2 more answers
For the sentence below, write an equivalent equation in symbols.
11111nata11111 [884]

Answer:

  x/3 +x/2 = x-3

Step-by-step explanation:

One-third of the number is x/3.

One-half of the number is x/2.

Three less than the number is x-3.

The expression you want is ...

  x/3 +x/2 = x-3

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The number is 18.

7 0
3 years ago
A triangle in the coordinate plane has the following vertices: P(-4,-3), 1(-2,-2), G(-2,-4). What are the new coordinates if the
Aloiza [94]

Answer:

P’(-1, 1), I’ (1, 2), G’(1,0)

Step-by-step explanation:

7 0
2 years ago
Which values are within the range of the piecewise-defined function?
musickatia [10]

Answer:  y < 1

<u>Step-by-step explanation:</u>

\begin{array}{c|c||l}\underline{x; x-3}&\underline{y=-x-2}&\\-3&-(-3)-2=1&\text{approaching but not including 1}\\-2&-(-2)-2=0&\\-1&-(-1)-2=-1&\end{array}

The first function has the range of y < -4

The second function has the range of y = -3

The third function has the range of y < 1

The largest y-value is 1 and the smallest y-value is -∞, therefore the range (y-values) are from -∞ to 1 → y < 1

6 0
3 years ago
Read 2 more answers
The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f
Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

h(t)=-16t^{2}+25t+5

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

h(t)=-16(t^{2}-\frac{25}{16}t)+5

Complete the square

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}

Rewrite as perfect squares

h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

The vertex is the point (\frac{25}{32},\frac{945}{64})

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

16(t-\frac{25}{32})^{2}=\frac{945}{64}

(t-\frac{25}{32})^{2}=\frac{945}{1,024}

square root both sides

(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}

t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}

the positive value is

t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec

8 0
3 years ago
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